Amixture of nitrogen and neon gas is compressed from a volume of to a volume of , while the pressure is held constant at . calculate the work done on the gas mixture. round your answer to significant digits, and be sure it has the correct sign (positive or negative).

-266 kJ

Explanation:

There is some info missing. I think this is the original question.

A mixture of nitrogen and neon gas is expanded from a volume of 53.0 L to a volume of 90.0 L, while the pressure is held constant at 71.0 atm. Calculate the work done on the gas mixture. Be sure your answer has the correct sign (positive or negative) and the correct number of significant digits.

Given data

Initial volume = 53.0 L

Final volume = 90.0 L

Pressure = 71.0 atm

We can find the work (w) associated with the expansion of the gaseous mixture using the following expression.

w = - P × ΔV = - 71.0 atm × (90.0 L - 53.0 L) = -2.63 × 10³ atm . L

w = -2.63 × 10³ atm . L × (101.3 J/ 1 atm . L) = -2.66 × 10⁵ J = -266 kJ

The negative sign means that the system does work on the surroundings.

im not exactly sure but i think its b!

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.